7. Find the percentage decrease in weight of a body when taken 16 km below the surface of the Earth. What happens to the weight of the body when it is taken to the center of the Earth? Given \( R_{e}=6400 \mathrm{~km} \).

For an object at the Earth’s surface, the gravitational acceleration is given by   g = GM/R² When the object is at a depth d (here, d = 16 km), its distance from the Earth’s center becomes   r = R – d Assuming that the Earth has uniform density, the gravitational acceleration inside the Earth varies linearly with distance from the center. Thus, at depth d the gravitational acceleration is   g' = g × (r/R) = g × ((R – d)/R) For R = 6400 km and d = 16 km, we have   g' = g × ((6400 – 16)/6400) = g × (6384/6400) Calculating the fraction:   6384/6400 = 1 – (16/6400) = 1 – 0.0025 = 0.9975 So, the weight (which is proportional to the gravitational acceleration) decreases by   (1 – 0.9975) × 100% = 0.25% Thus, when the object is taken 16 km below the surface, its weight decreases by approximately 0.25%. Now, when the object is taken to the center of the Earth, r = 0. Since the gravitational acceleration inside is given by   g' = g × (r/R) at the center, g' = 0. This means that the object experiences no net gravitational force at the center—its weight effectively becomes zero. In summary: 1. The weight decreases by approximately 0.25% when taken 16 km below the surface. 2. At the center of the Earth, the weight becomes zero.